UNIPLANAR MOTION OF A RIGID BODY 225 where s is the length of the string which is unwound. Substituting these in the energy equation we obtain , or = co0~ + -y- S = co02+270, (a = 00) which is the equation (3) obtained by the torque method. Differentiating 'the last equation we have 2 co = 2 700 at :and dco = 7, which is the equation obtained by the torque method; therefore the rest of the problem is identical with that given by the torque method. 2. A flywheel rotates about a horizontal axis under the action of a falling body, which is suspended by means of a string wound around the axle of the flywheel. Discuss the motion, neglecting the mass of the string. Let I = the moment of inertia of the rotating system. m = the mass of the falling body. a = the radius of the axle. T = the tensile force of the string. TOKQUE METHOD. Taking the moments about the axis of rotation we have Q = Ta for the resultant torque. There-f ore Icb = Ta is the torque equation. But considering the forces acting upon the falling body we get mv = mg T. Hence 7cb = Ta = (mg mv) a = m (g acib) a. FIG. 112. ENERGY METHOD. Suppose the flywheel to start from rest, and let h denote the distance covered by the body during its fall. Then the energy equation gives J Jco2 + £ mv2 = mgh. Differentiating with respect to t, luu + mvv = mgh. But v = au and h = v, therefore /coco + wa2cocb = mgaco, or Jco = m (g acb) a.